\section{Polar Decomposition}

As we know, a square matrix \textbf{A} can be decomposed into two parts, a symetric semi-definite part \textbf{S} and a rotational part \textbf{R}, which should look like the following,

\begin{equation}\nonumber
\textbf{A} = \textbf{RS}
\end{equation}

It's quite easy to show how it works from scratch,

\begin{equation}\nonumber
\begin{split}
\textbf{A} = \textbf{U}\boldsymbol{\Sigma}\textbf{V}^{T} = (\textbf{UV}^{T})(\textbf{V}\boldsymbol{\Sigma}\textbf{V}^{T}) = \textbf{RS}
\end{split}
\end{equation}

Here notice \textbf{U} and \textbf{V} are assumed to be orthogonal but not rotational, as reflection could also occur by doing SVD. However, once we get the SVD, we can always move the negative part into the singular matrix. Now we start our problem. The rotational part is actually to minimize the following objective,

\begin{equation}\nonumber
\begin{split}
\min&\quad \|\textbf{A} - \textbf{R}\|^{2}_{F}\\
s.t.&\quad \textbf{RR}^{T} = \textbf{I}\\
&\quad det(\textbf{R}) = 1
\end{split}
\end{equation}

Now I will give the proof on how the above SVD formula satisfies the optimal solution.

\begin{equation}\nonumber
\begin{split}
\|\textbf{A} - \textbf{R}\|^{2}_{F} &= tr[(\textbf{A} - \textbf{R})^{T}(\textbf{A} - \textbf{R})] = tr[\textbf{A}^T\textbf{A} - \textbf{A}^T\textbf{R} - \textbf{R}^T\textbf{A} + \textbf{I}]\\
&= tr[\textbf{A}^T\textbf{A} - 2\textbf{R}^T\textbf{A} + \textbf{I}]\\
\end{split}
\end{equation}

So far, it's quite clear the minimization problem is actually equivalent to the following maximization problem,

\begin{equation}\nonumber
\begin{split}
\max&\quad \textbf{R} : \textbf{A}\\
s.t.&\quad \textbf{RR}^{T} = \textbf{I}\\
&\quad det(\textbf{R}) = 1
\end{split}
\end{equation}

which can be shown as following,

\begin{equation}\nonumber
\begin{split}
\textbf{R} : \textbf{A} = tr[\textbf{V}^{T}\textbf{R}^T\textbf{U}\boldsymbol{\Sigma}] = tr[\textbf{Q}\boldsymbol{\Sigma}]
\end{split}
\end{equation}

Here \textbf{Q} is also an orthogonal matrix, which is simply $\textbf{V}^{T}\textbf{R}^T\textbf{U}$. Then the problem becomes,

\begin{equation}\nonumber
\begin{split}
\max&\quad tr[\textbf{Q}\boldsymbol{\Sigma}]\\
s.t.&\quad \textbf{QQ}^{T} = \textbf{I}
\end{split}
\end{equation}

Here notice all the singular values are non-negative, thus the maximum of the trace one can get is the sum of these singular values. It further implies that \textbf{Q} is identity matrix. Then we have $\textbf{R} = \textbf{UV}^T$.